Gautschi's inequality

There are several known proofs of Gautschi's inequality. One simple proof is based on the strict logarithmic convexity of Euler's gamma function. By definition, this means that for every ${\displaystyle u}$ and ${\displaystyle v}$ with ${\displaystyle u\neq v}$ and every ${\displaystyle t\in (0,1)}$, we have

${\displaystyle \Gamma (tu+(1-t)v)<\Gamma (u)^{t}\Gamma (v)^{1-t}.}$

Apply this inequality with ${\displaystyle u=x}$, ${\displaystyle v=x+1}$, and ${\displaystyle t=1-s}$. Also apply it with ${\displaystyle u=x+s}$, ${\displaystyle v=x+s+1}$, and ${\displaystyle t=s}$. The resulting inequalities are:

${\displaystyle {\begin{aligned}\Gamma (x+s)&<\Gamma (x)^{1-s}\Gamma (x+1)^{s}=x^{s-1}\Gamma (x+1),\\\Gamma (x+1)&<\Gamma (x+s)^{s}\Gamma (x+s+1)^{1-s}=(x+s)^{1-s}\Gamma (x+s).\end{aligned}}}$

Rearranging the first of these gives the lower bound, while rearranging the second and applying the trivial estimate ${\displaystyle x+s<x+1}$ gives the upper bound.

A survey of inequalities for ratios of gamma functions was written by Qi.^[3]

The proof by logarithmic convexity gives the stronger upper bound

${\displaystyle {\frac {\Gamma (x+1)}{\Gamma (x+s)}}<(x+s)^{1-s}.}$

Gautschi's original paper proved a different, stronger upper bound,

${\displaystyle {\frac {\Gamma (x+1)}{\Gamma (x+s)}}\leq \exp((1-s)\psi (x+1)),}$

where ${\displaystyle \psi }$ is the digamma function. Neither of these upper bounds is always stronger than the other.^[4]

Kershaw proved two tighter inequalities. Again assuming that ${\displaystyle x>0}$ and ${\displaystyle s\in (0,1)}$,^[5]

${\displaystyle {\begin{aligned}\left(x+{\frac {s}{2}}\right)^{1-s}&<{\frac {\Gamma (x+1)}{\Gamma (x+s)}}<\left[x-{\frac {1}{2}}+\left(s+{\frac {1}{4}}\right)^{1/2}\right]^{1-s},\\\exp \left((1-s)\psi (x+s^{1/2})\right)&<{\frac {\Gamma (x+1)}{\Gamma (x+s)}}<\exp \left((1-s)\psi \left(x+{\frac {1}{2}}(s+1)\right)\right).\end{aligned}}}$

Gautschi's inequality is specific to a quotient of gamma functions evaluated at two real numbers having a small difference. However, there are extensions to other situations. If ${\displaystyle x}$ and ${\displaystyle y}$ are positive real numbers, then the convexity of ${\displaystyle \psi }$ leads to the inequality:^[6]

${\displaystyle {\frac {1}{2}}(\psi (x)+\psi (y))\leq {\frac {\log \Gamma (y)-\log \Gamma (x)}{y-x}}\leq \psi \left({\frac {x+y}{2}}\right).}$

For ${\displaystyle s\in (0,1)}$, this leads to the estimates

${\displaystyle \exp {\bigl (}(1-s)\psi (x+s){\bigr )}\leq {\frac {\Gamma (x+1)}{\Gamma (x+s)}}\leq \exp \left((1-s)\psi \left(x+{\frac {1}{2}}(s+1)\right)\right).}$

A related but weaker inequality can be easily derived from the mean value theorem and the monotonicity of ${\displaystyle \psi }$.^[7]

A more explicit inequality valid for a wider class of arguments is due to Kečkić and Vasić, who proved that if ${\displaystyle y>x>1}$, then:^[8]

${\displaystyle {\frac {y^{y-1}}{x^{x-1}}}e^{x-y}<{\frac {\Gamma (y)}{\Gamma (x)}}<{\frac {y^{y-1/2}}{x^{x-1/2}}}e^{x-y}.}$

In particular, for ${\displaystyle s\in (0,1)}$, we have:

${\displaystyle {\frac {(x+1)^{x}}{(x+s)^{x+s-1}}}e^{-(1-s)}<{\frac {\Gamma (x+1)}{\Gamma (x+s)}}<{\frac {(x+1)^{x+1/2}}{(x+s)^{x+s-1/2}}}e^{-(1-s)}.}$

Guo, Qi, and Srivastava proved a similar-looking inequality, valid for all ${\displaystyle y>x>0}$:^[9]

${\displaystyle {\frac {(x+1)^{x+1}}{(y+1)^{y+1}}}e^{y-x}<{\frac {\Gamma (x+1)}{\Gamma (y+1)}}<{\frac {(x+1/2)^{x+1/2}}{(y+1/2)^{y+1/2}}}e^{y-x}.}$

For ${\displaystyle s\in (0,1)}$, this leads to:

${\displaystyle {\frac {(x+1)^{x+1}}{(x+s)^{x+s}}}e^{s-1}<{\frac {\Gamma (x+1)}{\Gamma (x+s)}}<{\frac {(x+1/2)^{x+1/2}}{(x+s-1/2)^{x+s-1/2}}}e^{s-1}.}$

• Gautschi Walter, (1959), Some Elementary Inequalities Relating to the Gamma and Incomplete Gamma Function, Journal of Mathematics and Physics, 38, doi:10.1002/sapm195938177.